If #k_f=1.86*""^(@)C*mol^-1*kg# for water, what is the freezing point of a solution composed of #255*kg# #CaCl_2#, and #33.8*L# of water?

1 Answer
anor277
Apr 27, 2017

Well, #DeltaT_"fusion"=k_fxxcxxi# where.................

Explanation:

Where #k_f# is the #"molal freezing point depression constant"#, which you have kindly given, #c="molal concentration"# in #mol*kg^-1#, and #i#, the so-called #"van't Hoff factor"# which reflects the speciation of an ionic solid in a solvent.

Clearly, here, #i=3#. Why? Because in solution, #"calcium chloride"# speciates to give three ions......

#CaCl_2(aq) stackrel(H_2O)rarrCa^(2+) + 2Cl^(-)#

And so...............

#Delta_"fusion"=1.86*""^(@)C*m^-1xx3xx((2.55xx10^3*g)/(110.98*g*mol^-1))/(33.8*kg)=3.80# #""^@C#.

This reflects the depression of the fusion point (the melting point) with RESPECT to PURE SOLVENT. And thus #"fusion point"# #-=# #-3.80# #""^@C#.

In practice, the melting point probably would not be depressed so far. And this is the province of measurement.

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